Er, yes, sorry. And even if my ODE had been right, my subsequent statement about arc length would have been wrong, as I accidentally wrote down the arc length in a Cartesian (u,v) plane rather than that on the sphere!
> Clive Tooth went on to write in another article: > > > dv/du = k cos v > > Anyway, this is the way I see it... > u is the longitude and v is the latitude. > Let r be the radius of the Earth. > Let a be the angle made by the loxodrome with each parallel of latitude. > Let tan a = k. > Let the initial and final latitudes be Lat0 and Lat1. > Let the initial and final longitudes be Lon0 and Lon1. > Let the latitude and longitude of M be LatM and LonM. > > Now, > ds/dv = r cosec a > > Integrating from P0 to P1 gives: > > s = (Lat1-Lat0) r cosec a > > In other words, the distance is _linear_ in the latitude. > So > LatM = (Lat0+Lat1)/2 (1)
Yes. I realised this on the way home last night, but have no computer at home so I couldn't post a correction.
> Again, > k = sec v dv/du (2) > > Integrating from P0 to P1 gives: > > k(Lon1-Lon0) = log((sec Lat1 + tan Lat1)/(sec Lat0 + tan Lat0)) > > So > k = log((sec Lat1 + tan Lat1)/(sec Lat0 + tan Lat0))/(Lon1-Lon0) > (3)
> Integrating (2) from P0 to M gives: > > k(LonM-Lon0) = log((sec LatM + tan Lat1)/(sec LatM + tan Lat0))
Where do these expressions
sec (one angle) + tan (a different angle)
come from? I get
k(LonM-Lon0) = log((sec LatM + tan LatM)/(sec Lat0 + tan Lat0)).
However I haven't checked the ways you can transform this using (1), so it may be equivalent to yours.
> giving > > LonM = Lon0+log((sec LatM + tan Lat1)/(sec LatM + tan Lat0))/k > (4) > > Where LatM and LonM are already know from (1) and (3) above.
I don't see why (4)'s right (see comment above).
I'd like to get a symmetric expression for LonM. To reduce the clutter, let's write f(x) = sec x + tan x. (As far as I can see, we gain nothing from the fact that f(x) can also be expressed as tan (pi/4 + x/2) or whatever it is.) Then the equation of the curve is